A satellite of mass \(1000 ~\text{kg}\) is launched to revolve around the earth in an orbit at a height of \(270 ~\text {km}\) from the earth's surface. Kinetic energy of the satellite in this orbit is: (in \(\times~ 10^{10}~\text {J}\))​​​​​​ (Mass of earth \(= 6 \times 10^{24}~\text {kg}, \) Radius of earth \(= 6.4 \times 10^6 ~\text m, \) Gravitational constant \(= 6.67 × 10^{-11} ~\text{Nm}^2\text{ kg}^{-2}\))
1. \(4\) 
2. \(10\) 
3. \(3\)
4. \(14\)
Subtopic:  Orbital velocity |
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 A satellite of mass \(\dfrac{M}{2}\) is revolving around earth in a circular orbit at a height of \(\dfrac{R}{3}\) from earth surface. The angular momentum of the satellite is \(M \sqrt{\dfrac{G M R}{x}} .\). The value of \(x\) is: (where \(M,\) \(G\) and \(R\) are the mass, gravitational constant, and radius of Earth, respectively).
1. \(7\)
2. \(3\)
3. \(5\)
4. \(1\)

 
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The mass and radius of orbit for the two satellites are \((m,r)\) and \((3m, 3r)\) respectively. What will be the ratio of their orbital velocity about the Earth?
1. \(\sqrt 3 :1\)
2. \(1 : \sqrt 3 \)
3. \(\sqrt 2 :1\)
4. \(1:2\)
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A body is moving in a low circular orbit about a planet of mass \(M\) and radius \(R\). The radius of the orbit can be taken to be \(R\) itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is:
1. \(\sqrt{2}\)
2. \(\dfrac{1}{\sqrt{2}}\)
3. \(2\)
4. \(1\)

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A spacecraft orbits a planet at an altitude of \(20~\text{km}\) above its surface. Assuming that the planet’s gravitational force is the only force acting on the spacecraft, how many complete revolutions does it make in \(24\) hours? Given that the mass and radius of the planet are \(8 \times 10^{22} ~\text{kg}\) and \(2 \times 10^6 ~\text{m}\) respectively, and the gravitational constant is \(6.67 \times 10^{-11} ~\text{Nm}^2 / \text{kg}^2.\)
1. \(13\)
2. \(9\)
3. \(17\)
4. \(11\)

Subtopic:  Orbital velocity |
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